09P Oscillations NJC problem set 2025

1 09 Oscillation Problem Solutions P1 (a) Only for half the cycle (see notes) (b) Only for half the cycle (see notes) (c) No (always in opposite direction 𝑎 = −𝜔2𝑥) P2 D A – Amplitude is half of 70 cm B – Kinetic energy is max at equilibrium which occurs at T/4 C – Restoring force is maximum at amplitude and zero at equilibrium so restoring force decreases from t = 0 to t = T/4 D – Correct. Kinetic energy and speed highest at equilibrium. P3 A 𝐸 = 1 2 𝑚𝜔2𝑥02 Hence 𝐸 ∝ 𝑥02 New energy is ¾ E, hence new 𝑥0 = √3 2 𝑥 So change in amplitude = (1 − √3 2 ) 𝑥 = 0.134𝑥 P4 a) (i) The radian is the unit of measurement for angles and is the ratio of the arc length subtended to the radius of the circle. (a) (ii) For simple harmonic motion, angular frequency is 2𝜋 times the frequency of oscillations. (b) (i) Total energy = maximum gravitational potential energy from equilibrium position = 𝑚𝑔ℎ = 0.120 × 9.81 × 0.04 = 4.7 × 10−3𝐽 (b) (ii) Total energy = 1 2 𝑚𝜔2𝑥02 = 4.7 × 10−3 1 2 𝑚4𝜋2𝑓2𝑥02 = 4.7 × 10−3 𝑓 = √ 2×4.7×10−3 0.120×4×𝜋2×0.082 = 0.56 𝐻𝑧 (𝑥0 = 0.080 𝑚) P5 (i) At t1 and t2, x = 1.3m from mid tide level, ie. equilibrium 1.3 = 3.7 𝑠𝑖𝑛 𝜔𝑡 By sub 𝜔 = 2𝜋/𝑇 𝜔𝑡1 = 0.35901 rad and 𝜔𝑡2= 2.7826 rad t1 = 2606 s and t2 = 20195 s Hence ∆t = t2 – t1= 17589 s = 18 000 s (2 s.f.) OR 290 min (2 s.f.) (ii) Rate of rising tide and rate of falling tide is the gradient of tangent drawn at t1 and t2 respectively = 3.7𝜔 𝑐𝑜𝑠 𝜔𝑡1

2 = (3.7 × 2𝜋 45600) 𝑐𝑜𝑠 [( 2𝜋 45600) (2606)] = 4.8 × 10-4 m s-1 P6 D GPE should increase linearly with upwards displacement. The extension decreases with upwards displacement so, EPE should decrease. It is a curve as EPE = ½ kx2 KE is maximum at equilibrium and zero at amplitude positions. TE is constant. P7 A Total energy must be constant, also check value of KE is max at centre and equal to 1 2 𝑚𝑣2 = 1.8𝐽. EPE has negative value as it is set to be zero at the center. The spring has extension at the center so at the top where the extension, the EPE must be lower and hence a negative value. P8 (a) (i) t3, t7 (ii) t4, t8 (b) (i) 𝑓 = 4200 60 = 70𝐻𝑧 (ii) 𝑎0 = 𝜔2𝑥0 = 4𝜋2𝑓2𝑥0 = 4𝜋2 × 702 × 0.025 = 4800 𝑚𝑠−2 (c) P9 (a) (i) Simple harmonic motion occurs when the acceleration of an object is directly proportional to its displacement and the acceleration is always opposite in direction to the displacement. (ii)1 The graph shows that displacement can be both positive and negative, hence the mass is oscillating. (ii)2 The graph is not a straight line (curved at one end), hence acceleration is not directly proportional to displacement and the oscillations are not simple harmonic. (b) (i)1 At the upper amplitude position (i)2 When the sand first loses contact, the maximum acceleration is equal to the acceleration due to free fall 𝜔2𝑥0 = 𝑔 (2𝜋 × 13)2𝑥0 = 9.81 B1 straight line through the origin B1 negative gradient B1 correct values for amplitude (2.5 cm) and maximum acceleration acceleration displacement

3 𝑥0 = 1.47 𝑚𝑚 (ii) The minimum amplitude will not be different. The mass loses contact when the acceleration of the oscillations exceed the acceleration due to free fall. This is not dependent on mass and hence the size of the mass does not matter. (c) (i)1 𝐸𝑇 == 1 2 × 1.2 × (2𝜋 × 2.5)2 × 0.0342 = 0.171 𝐽 (i)2 When the potential energy is half the total energy, potential and kinetic energy are equal 𝐸𝑃 = 0.171 2 = 1 2 𝑚𝜔2𝑥2 𝑥 = 2.4 𝑐𝑚 (i)3 Total energy – straight line at 0.171 J from -3.4 cm to +3.4 cm Kinetic energy – inverted U shape curve. Kinetic energy is zero at amplitude and 0.171 J at zero displacement. Potential energy – U shape curve. 0.171 J at amplitude and zero at zero displacement. At 2.4 cm, Kinetic and potential energy graphs intercept at 0.086 J P10 (c) (i) Since ρ, A, g and m are all constants, the value of 𝜌𝐴𝑔 𝑚 is a constant. Hence a is directly proportional to x but in the opposite direction to the displacement from the equilibrium position as indicated by the negative sign. (c) (ii) Since it is in SHM, acceleration a = - ω2x This implies 𝜔2 = 𝜌𝐴𝑔 𝑚 𝑓 = 1 2𝜋 √𝜌𝐴𝑔 𝑚 𝑓 = 1 2𝜋 √(1000)(4.2 ×10−4)(9.81) 0.032 = 1.81 Hz (d) (i) Using Fig. 2 1. For 3 cycles, total time from graph = 1.50 s Period = 0.50 s f = 1/0.50 = 2.0 Hz 2. Since 𝜔2 = 𝜌𝐴𝑔 𝑚 only density changed. 4𝜋2𝑓2 = 𝜌𝐴𝑔 𝑚 𝜌 = 4𝜋2𝑓2𝑚 𝐴𝑔 = 4𝜋2(2.0)2(0.032) (4.2×10−4)(9.81) = 1226 kg m-3 (ii) 1. The decrease in amplitude shows that the total energy is decreasing with time due to light damping. This comes about due to (1) the viscous force between the liquid and the tube and (2) the resistive force between the sand and the tube.

4 2. Decrease in energy = (Total energy at t = 0) – (Total energy at t =1.0s) Since total energy = ½ mω2x02 Decrease in energy = ½ (0.032)(2π × 2)2[(1.5 × 10-2)2 – (0.85× 10-2)2] = 3.86 × 10-4 J P11 ai(1) Displacement is the distance moved from the equilibrium position. ai(2) Amplitude is the maximum displacement. aii) Simple harmonic motion occurs when the acceleration of an object is directly proportional to its displacement and the acceleration is always opposite in direction to the displacement. b) simple pendulum – the restoring force is the component of the weight tangential to the tension in the string towards the equilibrium position Floating block – the restoring force is provided by the resultant force due to upthrust and weight. Below the equilibrium position, the upthrust is greater than weight and results in an upwards net force which is also the restoring force. Above the equilibrium position, the weight is greater than upthrust resulting in a downwards net force which is also the restoring force. c) At equilibrium when the steel strip is depressed, the weight of mass M causes a clockwise moment about the clamp. A counterclockwise moment needs to be provided to ensure that the net moment about the clamp is zero. The upper part of part of the steel strip is stretched and in tension. The bottom part of the steel strip is compressed and provides a pushing force on the mass. Together, these forces provide a counterclockwise moment which is needed to result in the net moment for the block mass M to be zero. d) As C, E, L and M are constants, the equation can be expressed in the form a = -kx, showing that acceleration is directly proportional to displacement and opposite in direction to the displacement. Hence, the strip is undergoing simple harmonic motion. ei) for 4 oscillations, time is 0.84s, period T = 0.21s 𝜔 = 2𝜋 𝑇 = 29.9 𝑟𝑎𝑑 𝑠−1 eii) 𝐶𝐸 𝐿3𝑀 = 𝜔2 𝐶 = 𝜔2𝐿3𝑀 𝐸 = 3.44 × 10−10 7f) Since the frequency is the same, 𝜔 is the same L, C are the same, E is then proportional to M

5 𝑀 = 7.1×1010 2.0×1011 × 0.15 = 0.0533 𝑘𝑔 P12 (a) (i) 2 cycles = 1.2 s T = 0.6 s (ii) ω = 2π/T = 10.5 rad s-1 (b) (i) Damping is the phenomenon where the energy of the oscillating system decreases with time. (ii) 1. Light damping can be achieved by attaching a cardboard of large surface area to the base of the mass. This results in an increase in air resistance. OR Having the mass oscillate in a viscous fluid such as oil or honey. This results in an increase in viscous force between the fluid and the mass. 2. Increase the surface area of the cardboard OR viscosity of the fluid. P13 (a) Extension of spring is given by x =(l - 12cm). From the graph, when l = 12.0 cm, extension is zero and after l = 12.0 cm (which is the extension), the gradient of the graph is positive and constant showing that F is directly proportional to the extension (l – 12 cm). Hence the spring obeys Hooke’s law. (b) Work done = Area under the Force- extension graph = 0.5 × 1.4 × (18 × 10-2) = 0.126 J (c) (i) Max speed = ωx0 = (2π/0.84)